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That's because there is also a precondition for the domain to be bounded, and since the real numbers are not bounded, the theorem does not apply here (or for an
by arbitrarybits 3y ago
That's because there is also a precondition for the domain to be bounded, and since the real numbers are not bounded, the theorem does not apply here (or for any function R -> R).
- amalcon 3y agoYou can just do the same thing in the integers modulo any value greater than 1, creating a bounded domain without creating a fixed point. I imagine this runs afoul of a different precondition which I haven't identified, though -- it's obviously way more likely than that the entire field has managed to miss something this obvious. Edit: Actually it's obviously just that this set is not convex, or even continuous.